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I'm stuck on this question:

The original question is: find $\lim\limits_{x \to 0^+} x^{1/x}$.

This is equal to $\lim\limits_{x \to 0^+}e^{\frac{\ln(x)}{x}}$

I'm stuck at calculating $\lim\limits_{x \to 0^+}\frac{\ln(x)}{x}$.

I can't see how I can use L'Hopital's Rule or the Squeeze Theorem to solve this.

I know the final answer is $0$, so the limit in the exponent has to be negative infinity, but I don't understand why this is the answer.

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  • $\begingroup$ Lhopital's rule does not apply to your situation. Note that $\ln x\to-\infty$ as $x\to 0^+$, and $x\to 0$ as $x\to 0^+$, thus $\ln x /x\to-\infty$ as $x\to 0^+$. $\endgroup$ – Frank Lu Mar 16 '15 at 23:28
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There is no difficulty here :

$ \lim_{x \to 0} \; \ln(x) = -\infty $

and

$ \lim_{x \to 0} \; \dfrac{1}{x} = +\infty $

By product you have your $-\infty$. Used in exp you find the final result.

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  • $\begingroup$ sure its infinity * infinity.... silly me... $\endgroup$ – AK_ Mar 16 '15 at 23:37
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L'Hospital does not apply: as $x\to0^+$, we have $\log(x)\to-\infty$ and $x\to0^+$.

Ignoring the signs at first, what happens when the numerator gets big and the denominator gets small?

Next, apply the signs.

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